98=100 2 98*98*98= nbsp;983=(100 2)3 nbsp;(a b)3= nbsp;a3 b3 3a2b+3ab2= nbsp;1003 23 3(100)2(2)+3(100)22=1000000 8 60000+1200=941192 ...

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Sum of Consecutive Numbers (Not Starting from 1)

98*98*98

Question

989898

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Solution

98=100-2
$98 * 98 * 98 = 98^{3} = (100 - 2)^{3} (a - b)^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2} = 100^{3} - 2^{3} - 3 (100)^{2} (2) + 3 (100) 2^{2} = 1000000 - 8 - 60000 + 1200 = 941192$

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